Question

a parabola has a vertex with coordinates -4,1 and goes through the points coordinates 2,-17. find the equation of the parabola in the form of y=ax^2+bx+c?

Answer 1

hmm do you know the vertex form of a parabola?

Answer 2

some of it
i dont unde stand this question
if u an explain it im sue ill undestand

Answer 3

well if you notice at the figure http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

the one on the right side is the vertex form

so we know where the vertex is at, (-4, 1) that means the (h, k) are those values
h = -4 k= 1
thus \(\bf y = a(x-h)^2+k \implies y = a(x-(-4))^2+ 1\implies y = a(x+4)^2+ 1\)

Answer 4

so we just need "a"

so.. what's "a" anyway? well we dunno
but we know that another point, besides the vertex, in the parabola is (2, -17), since it goes through it

so let us use that ( 2, -17) meaning x = 2 y = -17

thus \(\bf y = a(x-h)^2+k \implies y = a(x-(-4))^2+ 1\implies y = a(x+4)^2+ 1\\ \quad \\
(2\quad ,\quad -17)\implies (-17) = a((2)+4)^2+ 1\)

solve for "a" to get "a"

Answer 5

once you get "a', to get the standard form just expand the binomial

Answer 6

k thanks for help

Answer 7

yw